Physics 115 Lecture 10

More about standing waves

February 13, 2018

1 Something cool

 Standing sound waves can be used to levitate and manipulate objects!

 Video 1 (D. Foresti – LTNT Switzerland)

 Video 2 (Y. Ochiai – Univ. Tokyo) Open tube air columns

 An antinode appears at each open end

 Higher harmonics follow a pattern similar to a string fixed at both ends

 Examples: organ pipe, flute Open tube air columns

 The effective length of the tube is actually longer than the physical tube. (See https://en.wikipedia.org/wiki/End_correction. For most homework and exam questions you do not need to know this correction. There is not agreement on a universal value; 0.46 works well in our lab.)

LLRtube 2  0.46  tube A standard concert flute has LR 0.610 m and 0.0095 m. v 343 m/s Ignoring end correction, f1    281 Hz 2Ltube 2 0.610 m Its effective length is L  0.610 m  2 0.46 0.0095 m  0.6188 m v 343 m/s f1, actual   277 Hz, a fourth of a step lower 2Ltube 2 0.6188 m

 Opening holes in the flute shortens its effective length Harmonic #antinodes #nodes λ f

fundamental 2L 2 1 f1 Chart of the harmonics 1

cfair  

2L second 3 2 2 f1 2

2L third 4 3 3f 3 1 What would happen to the fundamental wavelength of an organ pipe if you fill it with helium (c = 1025 m/s) instead of air (c = 343 m/s)?

A. The wavelength will increase. B. The wavelength will stay the same. C. The wavelength will decrease. What would happen to the fundamental wavelength of an organ pipe if you fill it with helium (c = 1025 m/s) instead of air (c = 343 m/s)?

A. The wavelength will increase. B. The wavelength will stay the same. C. The wavelength will decrease.

The wavelength is determined by the length of the pipe. What would happen to the fundamental frequency of an organ pipe if you fill it with helium (c = 1025 m/s) instead of air (c = 343 m/s)?

A. The frequency will increase. B. The frequency will stay the same. C. The frequency will decrease. What would happen to the fundamental frequency of an organ pipe if you fill it with helium (c = 1025 m/s) instead of air (c = 343 m/s)?

A. The frequency will increase. B. The frequency will stay the same. C. The frequency will decrease.

The speed increases while the wavelength remains the same: c = f λ Closed (at one end) tube air columns

 An antinode appears at the open end, node at the closed end

 Examples: clarinet, trumpet Closed (at one end) tube air columns

 The effective length of the tube is actually longer than the physical tube. (For most homework and exam questions you do not need to know this correction.)

 The end correction only applies to the open end, so the correction is smaller.

LLRtube 0.46  tube Harmonic #antinodes #nodes λ f

fundamental 4L 1 1 f1 Chart of the harmonics 1

cfair  

4L third 2 2 3 f1 3

4L fifth 3 3 5 f 5 1 The even harmonics are “missing” in a closed tube!

See: http://www.phys.unsw.edu.au/jw/flutes.v.clarinets.html What is the wavelength of the fifth harmonic standing wave in a 50-cm-long closed organ pipe?

A. 1.00 m B. 0.50 m C. 0.40 m D. 0.20 m What is the wavelength of the fifth harmonic standing wave in a 50-cm-long closed organ pipe?

A. 1.00 m 4L 4 0.50 m n    0.40 m B. 0.50 m n 5 This would correspond to C. 0.40 m c 343 m/s D. 0.20 m f    858 Hz  0.40 m You have an organ pipe that resonates at frequencies of 500, 700, and 900 Hz but nothing in- between. It may resonate at lower and higher frequencies as well. This pipe is A. open at both ends. B. closed at one end. C. either open or closed, it’s impossible to tell with the given information. You have an organ pipe that resonates at frequencies of 500, 700, and 900 Hz but nothing in- between. It may resonate at lower and higher frequencies as well. This pipe is A. open at both ends. B. closed at one end. TheC. harmonicseither openare separated or closed, by 200 it’s Hz. If this were an open pipe,impossible the lowest tofrequency tell with would the thus be 200 Hz. But thesegiven frequencies information. are odd; if you work backwards you find frequencies of 100, 300, 500, 700, 900, 1100 Hz, and so on. Because these are odd multiples of the lowest frequency, we conclude the pipe is closed at one end. In order to be tuned properly in a band, the funda- mental frequency of the bottle must be 440.0 Hz. If the bottle is 26.0 cm tall, how high h should it be filled with water? Assume c = 343 m/s. A. 4.4 cm B. 19.5 cm C. 39.0 cm D. 6.5 cm In order to be tuned properly in a band, the funda- mental frequency of the bottle must be 440.0 Hz. If the bottle is 26.0 cm tall, how high h should it be filled with water? Assume c = 343 m/s. A. 4.4 cm B. 19.5 cm C. 39.0 cm D. 6.5 cm 1 1c 1 343 m/s L    19.5 cm 4 4f 4 440.0 Hz h H  L 26.0 cm  19.5 cm  6.5 cm